Video: Powers in terms of Multiple Angles from Euler’s Formula

1. Use Euler’s formula to express sinΒ³ πœƒ cosΒ² πœƒ in the form π‘Ž sin πœƒ + 𝑏 sin 3πœƒ + 𝑐 sin 5πœƒ, where π‘Ž, 𝑏, and 𝑐 are constants to be found. 2. Hence, find the solutions of sin 5πœƒ βˆ’ sin 3πœƒ = 0 in the interval 0 β©½ πœƒ < πœ‹.

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Video Transcript

Use Euler’s formula to express sin cubed πœƒ cos squared πœƒ in the form π‘Ž sin πœƒ plus 𝑏 sin three πœƒ plus 𝑐 sin five πœƒ, where π‘Ž, 𝑏, and 𝑐 are constants to be found. Hence, find the solutions of sin five πœƒ minus sin three πœƒ equals zero in the interval πœƒ is greater than or equal to zero and less than πœ‹. Give your answer in exact form.

We begin by recalling the fact that sin πœƒ is equal to one over two 𝑖 times 𝑒 to the π‘–πœƒ minus 𝑒 to the negative π‘–πœƒ. And cos πœƒ is equal to a half times 𝑒 to the π‘–πœƒ plus 𝑒 to the negative π‘–πœƒ. This means we can find the product of sin cubed πœƒ and cos squared πœƒ. We can write it as one over two 𝑖 times 𝑒 to the π‘–πœƒ minus 𝑒 to the negative π‘–πœƒ cubed times a half times 𝑒 to the π‘–πœƒ plus 𝑒 to the negative π‘–πœƒ squared. One over two 𝑖 cubed is negative one over eight 𝑖. And a half squared is one-quarter. So we can rewrite our expression a little bit further. We find the product of negative one over eight 𝑖 and a quarter. And we get negative one over 32𝑖. And we can rewrite the rest of our expression as shown.

We’re now going to use the binomial theorem to expand each of the sets of parentheses. The first part becomes 𝑒 to three π‘–πœƒ plus three choose one 𝑒 to the two π‘–πœƒ times negative 𝑒 to the negative π‘–πœƒ and so on. And this simplifies to 𝑒 to the three π‘–πœƒ minus three 𝑒 to the π‘–πœƒ plus three 𝑒 to the negative π‘–πœƒ minus 𝑒 to the negative three π‘–πœƒ. Let’s repeat this process for 𝑒 to the π‘–πœƒ plus 𝑒 to the negative π‘–πœƒ squared. When we do, we get 𝑒 to the two π‘–πœƒ plus two 𝑒 to the zero which is just two plus 𝑒 to the negative two π‘–πœƒ.

We’re going to need to find the product of these two expressions. We’ll need to do that really carefully. We’ll need to ensure that each term in the first expression is multiplied by each term in the second expression. And we can write sin cubed πœƒ cos squared πœƒ as shown. Now, there’s quite a lot going on here. So you might wish to pause the video and double-check your answer against mine. We’re going to gather the corresponding powers of 𝑒 together.

We’ll gather 𝑒 to the five π‘–πœƒ and 𝑒 to the negative five π‘–πœƒ. We’ll collect 𝑒 to the plus and minus three π‘–πœƒ. And we’ll gather 𝑒 the π‘–πœƒ and 𝑒 to the negative π‘–πœƒ. Let’s neaten things up somewhat. We end up with negative one over 32𝑖 times 𝑒 to the five π‘–πœƒ minus 𝑒 to the negative five π‘–πœƒ minus 𝑒 to the three π‘–πœƒ minus 𝑒 to the negative three π‘–πœƒ minus two times 𝑒 to the π‘–πœƒ minus 𝑒 to the negative π‘–πœƒ. And now, you might be able to spot why we chose to do this. We can now go back to the given formulae. Let’s clear some space for the next step.

We kind of unfactorize a little. And we can rewrite sin cubed πœƒ cos squared πœƒ as shown. And we can therefore replace 𝑒 to the π‘–πœƒ plus 𝑒 to the negative π‘–πœƒ with sin πœƒ and so on. And we can see that sin cubed πœƒ cos squared πœƒ is equal to a 16th times two sin πœƒ plus sin three πœƒ minus sin five πœƒ. Since π‘Ž, 𝑏, and 𝑐 are constants to be found, we can say that π‘Ž, the coefficient of sin πœƒ, is one-eighth. 𝑏, the coefficient of sin three πœƒ, is a 16th. And 𝑐, the coefficient sin five πœƒ, is negative one 16th.

Let’s now consider part two of this question. We begin by using our answer to part one and multiplying both sides by 16. We then subtract two sin πœƒ from both sides and multiply through by negative one. And we can now see that we’ve got an equation in sin five πœƒ minus sin three πœƒ. We’re told that sin five πœƒ minus sin three πœƒ is equal to zero. So we let two sin πœƒ minus 16 sin cubed πœƒ cos squared πœƒ be equal to zero. And then, we factor by two sin πœƒ. Since the product of these two terms is equal to zero, this means that either of these terms must be equal to zero. So either two sin πœƒ is equal to zero and dividing by two, we could see that sin πœƒ is equal to zero or one minus eight sin squared πœƒ cos squared πœƒ is equal to zero. Given the interval πœƒ is greater than or equal to zero and less than πœ‹, we can see that one of our solutions is when πœƒ is equal to zero.

We’re going to rewrite our other equations somewhat. We know that sin two πœƒ is equal to two sin πœƒ cos πœƒ. Squaring this, we get sin squared two πœƒ equals four sin squared πœƒ cos squared πœƒ. And that’s in turn means that our equation is one minus two sin squared two πœƒ equals zero. Rearranging to make sin two πœƒ the subject, we see that sin two πœƒ is equal to plus or minus one over root two. Starting with the positive square root for πœƒ in the interval given, we know that sin two πœƒ is equal to one over root two when πœƒ is equal to πœ‹ by eight or three πœ‹ by eight. Similarly, we can solve for the negative square root. And we get five πœ‹ by eight and seven πœ‹ by eight. And there are therefore five solutions to the equation sin five πœƒ minus sin three πœƒ equals zero in the interval πœƒ is greater than or equal to zero and less than πœ‹. They are zero πœ‹ by eight, three πœ‹ by eight, five πœ‹ by eight, and seven πœ‹ by eight.

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